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# A certin deck of cards contains 2 blue cards,2 red cards,2

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Senior Manager
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A certin deck of cards contains 2 blue cards,2 red cards,2 [#permalink]  06 Jan 2004, 11:53
** A certin deck of cards contains 2 blue cards,2 red cards,2 yellow cards and 2 green cards.If two cards are randomly drawn from the deck,what is the probability that they both will not be blue?

1) 15/28
2) 1/4
3) 9/16
4) 1/32
5) 1/16

** A bag of 10 marbles contains 3 red marbles and 7 blue marbles.If 2 marbles are selected at randon ,what is the probability that at least one marble is blue?

1) 21/50
2) 3/13
3) 47/50
4) 14/15
5) 1/5

Vivek.
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"

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6/8 * 5/7 = 15/28

1- (Chance of red, red)
1- (3/10 * 2/9)

14/15

Will post reasoning later, but it ought to be pretty obvious...

(post edited for typographical error)

Last edited by stoolfi on 06 Jan 2004, 13:08, edited 1 time in total.
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stoolfi wrote:
6/8 * 5/7 = 15/28

1- (Chance of red, red)
1- (3/10 * 2/9)

13/15

Will post reasoning later, but it ought to be pretty obvious...

stoolfi - i'm sure you mean't 14/15. you're becoming like me.
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I agree with the first one. But wouldn't 2nd one be 14/15?

P of both red is 3/10 * 2*9 =6/90 = 1/15.
P of at least one blue is 1 - 1/15 = 14/15.
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drodgers wrote:
I agree with the first one. But wouldn't 2nd one be 14/15?

P of both red is 3/10 * 2*9 =6/90 = 1/15.
P of at least one blue is 1 - 1/15 = 14/15.

yes - that was a titleistesque stoolfi typo. i give him credit!
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stoolfi wrote:
6/8 * 5/7 = 15/28

1- (Chance of red, red)
1- (3/10 * 2/9)

14/15

Will post reasoning later, but it ought to be pretty obvious...

(post edited for typographical error)

This is how I had done it...

Both will not be blue= 1 - both of them are blue

= 1 - (2/8 * 1/7)
= 1 - (1/28)
= 15/28
Can anybody tell me where I have gone wrong.

I agree with the second answer.

Vivek.
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"

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vivek_dj wrote:
stoolfi wrote:
6/8 * 5/7 = 15/28

1- (Chance of red, red)
1- (3/10 * 2/9)

14/15

Will post reasoning later, but it ought to be pretty obvious...

(post edited for typographical error)

This is how I had done it...

Both will not be blue= 1 - both of them are blue

= 1 - (2/8 * 1/7)
= 1 - (1/28)
= 15/28
Can anybody tell me where I have gone wrong.

I agree with the second answer.

Vivek.

Probability of any color except blue (1) * probability of any color except blue (2)

6/8*5/7 =30/56 = 15/28

You just got there differently that's all
Senior Manager
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Sorry Titleist ..there was typo in my last post...

The answer I would get from my approach is 27/28.

I comprehend the solution given by you and stoolfi.

I just wanna know where am I going wrong in this approach.

Regards,
Vivek.
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"

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vivek, I see your point quite clearly. your answer is right since the question does specifically ask the probability of not getting two blue balls. since there's no answer choice for 27/28 I'm assuming that it's asking: "what's the probability that there are no blue balls chosen" where did you find this problem?
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Re: Probability Q's [#permalink]  07 Jan 2004, 03:15
Quote:
A certin deck of cards contains 2 blue cards,2 red cards,2 yellow cards and 2 green cards.If two cards are randomly drawn from the deck,what is the probability that they both will not be blue?

1) 15/28
2) 1/4
3) 9/16
4) 1/32
5) 1/16

6C2/8C2 = 15/28

Quote:
A bag of 10 marbles contains 3 red marbles and 7 blue marbles.If 2 marbles are selected at randon ,what is the probability that at least one marble is blue?

1) 21/50
2) 3/13
3) 47/50
4) 14/15
5) 1/5

1- 3C2 / 10C2 = 14/15
Re: Probability Q's   [#permalink] 07 Jan 2004, 03:15
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